# Decide whether or not $g$ is differentiable at $0$

I need some help with an analysis problem.

Define $$g:[-1,1]\to \Bbb R$$ by $$g(x)=(-1)^k/k^2$$ for $$|x|\in(1/(k+1),1/k], k=1,2,...$$ and $$g(0)=0$$. Decide whether or not $$g$$ is differentiable at $$0$$ and prove your answer.

What I have:

When $$k=1, |x|\in (1/2,1]$$ and if $$k=2,|x|\in(1/3,1/2]$$ so if $$k\to \infty, |x|\in(0,0]$$

But how would I find if this is differentiable at zero or not?

• Have you tried graphing this function? – Matthew Leingang Feb 19 '19 at 0:47
• I'm not quite sure how I would graph it. – geoplanted Feb 19 '19 at 0:50
• Good, so I think understanding what the function is doing well enough to graph it is a key step before deciding if it's differentiable at zero. What is $g(0.1)$? $g(0.01)$? $g(-0.1)$? Plot some points and get acquainted with $g$. – Matthew Leingang Feb 19 '19 at 0:52
• Thats something I am struggling with. If g is a function of x, how would I graph it since everything is defined in terms of k? – geoplanted Feb 19 '19 at 0:57
• From my understanding $g$ looks like a bunch of lines above and below the $x$ axis. The lines are getting shorter as they approach the origin. They also are getting closer together. For every $k$ there are two lines. The function is symmetrical. – ty. Feb 19 '19 at 1:05

## 1 Answer

We can look for the derivative by definition: $$g'(0) = \lim\limits_{h\to0}\frac{g(h)-g(0)}{h} = \lim\limits_{h\to0}\frac{g(h)}{h}$$

However, the way the function is defined is not very convenient for us to find the limit, because it is defined in terms of an arbitrary number $$k$$. That's why we will try to define the function in terms of $$x$$ only. Note that the condition $$|x| \in \left( \frac{1}{k+1}, \frac{1}{k} \right]$$ after playing around with inequalities (with $$x\neq 0$$), is equivalent with the condition $$\frac{1}{|x|} \in [k, k+1)$$ How can we express the function $$g(x)=(-1)^k/k^2$$ in terms $$x$$ only?

We can use the floor function. Note that the function $$\lfloor x \rfloor$$ is defined as the largest integer not less than $$x$$. With that in mind, it is easy to see that $$k=\lfloor 1/|x| \rfloor$$. Now the function can be written as $$g(x)=\frac{(-1)^{\lfloor 1/|x| \rfloor}}{\lfloor 1/|x| \rfloor^2}$$

Let's get back to the limit. Looking at the form of $$g(h)/h$$, we can intuit that this limit is zero. We shall prove this by definition. Namely, we must show that for all $$\epsilon>0$$, there exists $$\delta>0$$ such that $$0<|h|<\delta \implies \left|\frac{g(h)}{h}\right|<\epsilon$$

Since $$\left\lfloor \frac{1}{|h|} \right\rfloor \geq \frac{1}{2|h|} \implies \left\lfloor \frac{1}{|h|} \right\rfloor^2 \geq \frac{1}{4h^2}$$ for all sufficiently small $$h$$ (say for $$|h|<\eta$$), we have that

$$\left|\frac{g(h)}{h}\right| = \frac{1}{|h|\cdot\lfloor 1/|h| \rfloor^2} \leq \frac{4h^2}{|h|}=4|h|$$ For this to be smaller than $$\epsilon$$, it suffices to choose $$\delta<\min\{\epsilon/4,\eta\}$$.

Thus, we have proven that $$g'(0)=\lim\limits_{h \to 0} \frac{g(h)}{h}=0$$ and $$g$$ is differentiable at the point $$x=0$$. $$\ \ \rule{5pt}{5pt}$$

• Let me know if you need any more clarification. – Haris Gušić Feb 19 '19 at 2:29